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Prime polynomial

Posted by Chris on November 19, 2011 – 2:50 pm

How many positive integers n < 1000 makes the following be prime?:

3^n + 5^n + 7^n + 11^n + 13^n + 17^n + 19^n


This post is under “MathsChallenge” and has 6 respond so far.
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6 Responds so far- Add one»

  1. 1. slavy Said:

    Too easy, Chris :(

  2. 2. Chris Said:

    Hi Slavy. True, but it looks hard.

  3. 3. Chris Said:

    It looks so hard that nobody else is trying. In fact it’s a slightly trick question. The answer is 0 for any integer n > 0. Can anybody show why that’s the case?

  4. 4. jan Said:

    mod(3^n,3)=0

    n is odd
    mod(5^n,3)=0
    mod(7^n,3)=0
    mod(11^n,3)=0
    mod(13^n,3)=0
    mod(17^n,3)=0
    mod(19^n,3)=0

    n is even
    mod(5^n,3)=1
    mod(7^n,3)=1
    mod(11^n,3)=1
    mod(13^n,3)=1
    mod(17^n,3)=1
    mod(19^n,3)=1
    which adds to 6 and mod(6,3)=0

    so the number described is divisible by 3.

  5. 5. jan Said:

    mod(5^n,3)=-1
    mod(7^n,3)=1
    mod(11^n,3)=-1
    mod(13^n,3)=1
    mod(17^n,3)=-1
    mod(19^n,3)=1

    obv

  6. 6. Chris Said:

    Hi jan. That’s right. Thanks. I thought that I was going to have to answer it.

    For any n, 3^n + 5^n + 7^n + 11^n + 13^n + 17^n + 19^n
    = 0^n + (-1)^n + 1^n + (-1)^n + 1^n + (-1)^n + 1^n (mod 3)
    = 3*1^n + 3*(-1)^n = 0 (mod 3)
    So the “polynomial” is always divisible by 3 and can never be prime.

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